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MATHEMATICS (Issued by CENTRAL BOARD Of SECONDARY EDUCATION)

Below is the Question Paper Design in the subject of Mathematics (Class XII) for the Board Examination 2017

1. Question Paper Design

S.

No.

Typology Of Questions

VSA

(1 Mark)

SA

(2 Marks)

LA – I

(4 Marks)

LA – II

(6 Marks)

Marks Weightage

1 Remembering 2 2 2 1 20 20%

2 Understanding 1 3 4 2 35 35%

3 Application 1 - 3 2 25 25%

4 HOTS - 3 1 - 10 10%

5 Evaluation - - 1 (VBQ) 1 10 10%

Total 1 4 4 2 8 16 4 11 44 6 6 36 100 100%

2. Question-wise break-up

Type Of Questions

Marks per Question

Total no. of Questions

Total Marks

VSA 1 4 4

SA 2 8 16

LA – I 4 11 44

LA – II 6 6 36

Total 29 100

3. Difficulty level

Difficulty Level Weightage

Easy 20%

Average 60%

Difficult 20%

4. Choice(s)

There will be no overall choice in the question paper. However, 30% internal choices will be given in 4 marks and 6 marks questions.

Sample Paper Of CBSE (2017) By O. P. Gupta (INDIRA Award Winner)


Issued By CBSE For 2017 Examinations

For more stuffs on Maths, please visit : www.theOPGupta.com Time Allowed : 180 Minutes Max. Marks : 100

General Instructions : (a) All questions are compulsory. (b) The question paper consist of 29 questions divided into four sections A, B, C and D. Section A comprises of 04 questions of one mark each, section B comprises of 08 questions of two marks each, section C comprises of 11 questions of four marks each and section D comprises of 06 questions of six marks each. (c) All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. (d) There is no overall choice. However, internal choice has been provided in 03 questions of four marks each and 03 questions of six marks each. You have to attempt only one of the alternatives in all such questions.

SECTION A Question numbers 01 to 04 carry 1 mark each.

Q01. State the reason why the relation 2R {(a, b) : a b } on the set R of real numbers is not

reflexive. Q02. If A is a square matrix of order 3 and | 2A | k | A | , then find the value of k.

Q03. If a and b

are two non-zero vectors such that a b a .b

, then find the angle between a and b

.

Q04. If * is a binary operation on the set R of real numbers defined by a * b = a + b – 2, then find the identity element for the binary operation *.

SECTION B Question numbers 05 to 12 carry 2 marks each.

Q05. Simplify 1

2

1cot for x 1

x 1

.

Q06. Prove that the diagonal elements of a skew symmetric matrix are all zero.

Q07. If 1

2

5x 1 1y tan , x

1 6x 6 6

, then prove that 2 2

dy 2 3

dx 1 4x 1 9x

.

Q08. If x changes from 4 to 4.01, then find the approximate change in elog x .

Series : PTS/1 Code No. 16/8/1

Roll No.

Candidates must write the Code on the title page of the answer-book.

Please check that this question paper contains 4 printed pages. Code number given on the right hand side of the question paper should be written on the

title page of answer-book by the candidate. Please check that this question paper contains 29 questions. Please write down the Serial Number of the question before attempting it. 15 minutes time has been allotted to read this question paper. The question paper will be

distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the question paper only and will not write any answer on the answer-book during this period.

CBSE Sample Question Paper (2017)


Q09. Find 2

x

2

1 xe dx

1 x

.

Q10. Obtain the differential equation of the family of circles having centre at (0, b) and passing through the points (a, 0) and ( a,0) , where ‘b’ is the arbitrary constant.

Q11. If | a b | 60, | a b | 40 and | a | 22

, then find | b |

.

Q12. If 2 1 1

P(A) , P(B) , P(A B)5 3 5

, then find P(A | B) .

SECTION C Question numbers 13 to 23 carry 4 marks each.

Q13. If 1 2

A2 1

, then using 1A , solve the following system of equations :

x 2y 1, 2x y 2 .

Q14. Discuss the differentiability of the function

12x 1, x

12f (x) at x1 2

3 6x, x2

.

OR For what value of k, is

3 sin x cos x, x

6xf (x) 6

k, x6

continuous at x6

?

Q15. If x a sin pt, y bcospt , then show that 2

2 2 2

2

d y(a x )y b 0

dx .

Q16. Find the equation of the normal to the curve 22y x , which passes through the point (2, 1).

OR Separate the interval 0,2

into subintervals in which 4 4f (x) sin x cos x is strictly

increasing or strictly decreasing. Q17. A magazine seller has 500 subscribers and collects annual subscription charges of ` 300/- per

subscriber. She proposes to increase the annual subscription charges and it is believed that for every increase of ` 1/-, one subscriber will discontinue. What increase will bring maximum income to her? Make appropriate assumptions in order to apply derivatives to reach the solution. Write one important role of magazines in our lives.

Q18. Find 2 2

sin xdx

(cos x 1)(cos x 4) .

Q19. Find the general solution of the differential equation (1 tan y)(dx dy) 2xdy 0 .

OR Solve the differential equation x x

y y x1 e dx e 1 dy 0

y

.

Q20. Prove that a .{(b c) (a 2b 3c)} [a b c]

.

Q21. Find the values of a so that the lines x 1 y 2 z a x 4 y 1

; z2 3 4 5 2

are skew.

Q22. A bag contains 4 green and 6 white balls. Two balls are drawn one by one without replacement. If the second ball drawn is white, what is the probability that the first ball drawn is also white?

Q23. Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of diamond cards drawn. Also find the mean and the variance of the distribution.



SECTION D Question numbers 24 to 29 carry 6 marks each.

Q24. Let f :[0, ) R be a function defined by 2f (x) 9x 6x 5 . Prove that f is not invertible.

Modify, only the codomain of f to make f invertible and then find its inverse. OR Let * be a binary operation defined on Q Q by (a, b) * (c, d) = (ac, b + ad), where Q is

the set of rational numbers. Determine, whether * is commutative and associative. Find the identity element for * and the invertible elements of Q Q .

Q25. Using properties of determinants, prove that

2

23

2

(a b)c c

c

(b c)a a 2 a b c

a

(c a)b b

b

.

OR If p 0, q 0 and

p q p q

q r q r 0

p q q r 0

, then, using properties of determinants,

prove that at least one of the following statements is true : (a) p, q, r are in G.P.

(b) is a root of the equation 2px 2qx r 0 .

Q26. Using integration, find the area of the region bounded by the curves 2y 5 x and y x 1 .

Q27. Evaluate : 2

4 4

0

x sin x cos xdx

sin x cos x

.

OR Evaluate 4

2x

0

(x e )dx as the limit of a sum.

Q28. Find the equation of the plane through the point (4, –3, 2) and perpendicular to the line of intersection of the planes x y 2z 3 and 2x y 3z 0 . Find the point of intersection of

the line ˆ ˆ ˆ ˆ ˆ ˆr i 2 j k (i 3j 9k)

and the plane obtained above.

Q29. In a mid-day meal programme, an NGO wants to provide vitamin rich diet to the students of an MCD school. The dietician of the NGO wishes to mix two types of food in such a way that vitamin contents of the mixture contains at least 8 units of vitamin A and 10 units of vitamin C. Food 1 contains 2 units per kg of vitamin A and 1 unit per kg of vitamin C. Food 2 contains 1 unit per kg of vitamin A and 2 units per kg of vitamin C. It costs ` 50/- per kg to purchase Food 1 and ` 70/- per kg to purchase Food 2. Formulate the problem as LPP and solve it graphically for the minimum cost of such a mixture.



SOLUTIONS & MARKING SCHEME

SECTION A

Q01. As 2

1 1 1 1, R

2 2 2 2

. Hence R is not reflexive. [1]

Q02. As | 2A | k | A | 32 | A | k | A | k 8 . [1]

Q03. As a b a.b

a b a.b

ˆa b sin n a b cos

ˆa b sin n a b cos

sin 1 cos

otan 1 45 . [1] Q04. Let e be the identity element. Then a *e a e*a . So, a *e a a e 2 a e 2 . [1]

SECTION B

Q05. Let 1

2

1y cot , x 1

x 1

.

Put 1x sec sec x . Then sec 12

[1/2]

1 1

2

1 1y cot cot

tansec 1

1 11

y cot cot cottan

[1/2]

1y cot cot( ) 1y sec x . [1]

Q06. Let A be a skew-symmetric matrix. Then by def., TA A [1/2]

So, th T ththe (i, j) element of A the (i, j) element of ( A) [1/2]

th ththe (j, i) element of A the (i, j) element of A [1/2]

For the diagonal elements, th thi j then, the (i, i) element of A the (i, i) element of A

th2 the (i, i) element of A 0 ththe (i, i) element of A 0 [1/2]

Hence the diagonal elements of a skew symmetric matrix are all zero. # See a simple approach for this question in O.P. Gupta’s Mathematicia Vol. 1.

Q07. Here 1 1 1 1

2

5x 3x 2xy tan tan tan 3x tan 2x

1 6x 1 3x.2x

[1]

1 1dy dtan 3x tan 2x

dx dx

2 2

dy 3 2

dx 1 9x 1 4x

. [1]

Q08. Let ey log x, x 4, x 0.01 [1/2]

dy 1

dx x

1dy dx

x

x 4

dy 1dy x 0.01

dx 4

[1/2]

So, 1

dy 0.0025400

. [1]

Q09. Let 2

x

2

1 xI e dx

1 x

2x

2 2

(1 x)I e dx

(1 x )

2

x

2 2

1 x 2xI e dx

(1 x )

x

2 2 2

1 2xI e dx

1 x (1 x )

[1]

x

2

1I e C

1 x

as

2 2 2

d 1 2x

dx 1 x (1 x )

x xf (x) f (x) e dx f (x)e C . [1]

Q10. As 2 2 2x h y k r is the circle with centre at (h, k) and radius of r.



So, 2 2 2x 0 y b r , where

2 22 2 2r a 0 0 b a b

2 2 2 2 2x y 2by b a b 2 2 2x y 2by a …(i) [1]

dy dy

2x 2y 2b 0dx dx

x yy

by

[1/2]

Substituting value of b in (i), 2 2 2x yyx y 2 y a

y

2 2 2 2y (x y ) 2xy 2y y a y 2 2 2 2y (x y a 2y ) 2xy 0

That is, 2 2 2 dy(x y a ) 2xy 0

dx is the required differential equation. [1/2]

Q11. As 2 2 2 2| a b | | a b | 2 | a | | b |

[1]

2 2 2 260 40 2 22 | b |

2| b | 2116

[1/2]

Hence | b | 46

. [1/2]

Q12. As P(A B)

P(A | B)P(B)

[1/2]

1 P(A B)

1 P(B)

[1/2]

1 P(A) P(B) P(A B)

11

3

[1/2]

2 1 11

75 3 52 10

3

. [1/2]

SECTION C

Q13. Here 1 2

A2 1

Clearly, |A| = 5, 1 2

adj.A2 1

1 1 2adj.A 1A

2 1| A | 5

[1/2+(1+1/2)+1/2]

Given equations are x 2y 1, 2x y 2 .

Let 1 2 1 x

AX B where A , B , X2 1 2 y

[1/2]

So, 1 1A AX A B 1I X A B

1 2 11

X2 1 25

3x 5

y 4

5

[1/2]

By equality of matrices, we get : x 3/5, y 4/5 . [1/2]

Q14. Note that 1

f 1/2 3 6 02

Now h 0 h 0

11 12 h 1 0f h f

22 2Lf 1/2 lim lim 2

h h

[1+1/2]



And, h 0 h 0

11 13 6 h 0f h f

22 2Rf 1/2 lim lim 6

h h

[1+1/2]

Since Rf 1/2 Lf 1/2 so, f isn’t differentiable at x = 1/2. [1]

OR Here f k6

. [1/2]

Also, x

6

3 sin x cos xlim

x6

x 0

6

2sin x6

lim 2 1 2

x6

[2+1]

For the continuity of f (x) at x6

, we have

x 06

f lim f (x)6

k 2 . [1/2]

Q15. Here x a sin pt, y bcospt

Differentiating w.r.t. t,

dx dy

apcos pt, bpsin ptdt dt

dy b

tan ptdx a

[1+1/2]

On differentiating w.r.t. x, we get : 2 2

2

d y bpsec pt dt

dx a dx [1]

2 2

2 2 2

d y bpsec pt 1 b 1 1

dx a apcos pt a cos pt cos pt

2

2 2 2

d y b 1 b

dx a 1 sin pt y

2

22 2

2

d y b 1 b

xdx a y1

a

2

2 2 2

d y b b

dx a x y

22 2 2

2

d y(a x )y b

dx

2

2 2 2

2

d y(a x )y b 0

dx . [1+1/2]

Q16. Let the normal be at 1 1(x , y ) to the curve 22y x 21 12y x ...(a) [1/2]

Also dy

xdx

slope of normal at 1 1

1

1(x , y )

x . [1]

Equation of normal : 1 1

1

1y y (x x )

x

…(i) [1/2]

As (i) passes through (2, 1) so, 1 1 1 1

1

11 y (2 x ) x y 2...(ii)

x

[1/2]

Solving (a) and (ii), we get 2/3 1/31 1x 2 , y 2 [1/2]

Hence the required equation is 1/3 2/3

2/3

1y 2 (x 2 )

2

i.e., 2/3 2/3x 2 y 2 2 . [1]

OR 4 4Given f (x) sin x cos x, x 0,2

3 3f (x) 4sin x cos x 4cos x sin x

3 3f (x) 4sin x cos x 4cos x sin x 22f (x) 4sin x cos x(cos x sin x)

f (x) 2sin 2x cos 2x sin 4x [1]

For critical points, f (x) sin 4x 0 sin 4x 0 4x 0, , 2 , 3 , 4 ,....



x 0, , 0,4 2 2

[1]

Q17. Let the seller increases ` x in annual subscription.

Total revenue generated by her, R x ` 300 x 500 x [1+1/2]

Diff. w.r.t. x both sides, R x 300 x 500 x 200 2x [1/2]

Again diff. w.r.t. x, R x 2 [1/2]

For the points of local maxima & local minima, R x 0 x 100

Now, R x 100 2 0 R x is maximum at x 100 .

So, an increment of `100 will maximize the profit. [1/2] Important role of magazines in our lives : Magazines contribute a great deal, to the

development of our knowledge. Through valuable and subtle critical and commentary articles on culture, social civilization, new life style we learn a lot of interesting things. Through reading magazines, our mind and point of view are consolidated and enriched. [1]

Q18. Let 2 2

sin xI dx

(cos x 1)(cos x 4)

Put cos x t sin xdx dt so, 2 2

dtI

(t 1)(t 4)

[1]

Consider 2

2 2

1 1 A Bwhere y t

(t 1)(t 4) (1 y)(4 y) 1 y 4 y

[1/2]

1 A(4 y) B(1 y) 1 1

A , B3 3

[1]

2 2

1 1 1Therefore, I dt

3 t 4 t 1

1 11 1 t

I tan tan t C3 2 2

[1]

1 11 cos x 1I tan tan cos x C

6 2 3

. [1/2]

Q19. Here (1 tan y)(dx dy) 2xdy 0 (1 tan y)dx (1 tan y)dy 2xdy 0

dx 1 tan y 2x

0dy 1 tan y

dx 2x 1

dy 1 tan y

[1]

Clearly it is in the form of dx

P y x Q ydy

where 2

P y , Q y 11 tan y

sin y cosy2 (cos y sin y) ( sin y cos y)1 dydy dyP(y)dy cos y sin y1 tan y cosy sin yI.F. e e e e

y log(cos y sin y) y log(cos y sin y) ye e e e cos y sin y [2]

Therefore, the solution is given by x I.F. I.F. Q y dy C

y yxe cos y sin y e cos y sin y 1dy C

y yxe cos y sin y e sin y C . [1]

Interval Sign of f (x) f (x) is strictly

0,4

Negative Decreasing

,4 2

Positive Increasing

[1] [1]



OR We have x x

y y x1 e dx e 1 dy 0

y

x x

y y x1 e dx e 1 dy

y

x

y

x

y

xe 1

ydx xf

dy y1 e

say

Put x x, y y then,

x x

y y

x x

y y

x xe 1 e 1

y yx xf f

y y1 e 1 e

Hence the diff. eq. is homogeneous. [1/2]

To solve, let dx dv

x vy v ydy dy

[1/2]

v

v

e v 1dvv y

dy 1 e

v v v v

v v

dv ve e v ve e vy

dy 1 e 1 e

[1]

v

v

1 e dydv

v e y

vlog v e log y log C [1]

vlog (v e )y log C v(v e )y C [1/2]

x/ yxe y A, where A C

y

. [1/2]

Q20. LHS : a .{(b c) (a 2b 3c)} a .{b a 2b b 3b c c a 2c b 3c c}

[1]

a .{b a 0 3b c c a 2c b 0}

[1]

a .b a 3(a .b c) a .c a 2 (a .c b)

[1]

[a b a] 3[a b c] [a c a] 2[a c b]

[1/2]

0 3[a b c] 0 2[a b c] [a b c] RHS

. [1/2]

Q21. Let 1 2

x 1 y 2 z a x 4 y 1L : and; L : z

2 3 4 5 2

The d.r.’s of lines L1 and L2 are respectively 2, 3, 4 and 5, 2, 1 As 2 :3: 4 5 : 2 :1, the lines are not parallel . [1/2]

Now coordinates of any random point on the lines L1 and L2 are respectively

2 1,3 2, a and 5 4, 2 1, . [1+1]

Lines will be skew if, apart from being non-parallel, they do not intersect each other. So, there must not exist a pair of values of and which satisfy the following three equations

simultaneously : 2 1 5 4, 3 2 2 1, 4 a

Solving the first two equations, we get : 1 . [1]

If these values of and satisfy the third equation then, the lines will be intersecting each

other and hence won’t be skew lines. So, for lines L1 & L2 to be skew, we must have 4 a i.e., 4 ( 1) a 1 a 3 . [1/2]

Q22. Let 1E : first ball drawn is white , 2E : first ball drawn is green , A : second ball drawn is white .

1 2 1 2

6 4 5 6P(E ) , P(E ) , P(A | E ) and P(A | E )

10 10 9 9 . [1+1/2]



By Bayes’ theorem, 1 11

1 1 2 2

P(E )P(A | E )P(E | A)

P(E )P(A | E ) P(E )P(A | E )

[1]

6 5510 9

6 5 4 6 910 9 10 9

. [1+1/2]

Q23. Let X denote the random variable for the diamonds cards X 0,1, 2 .

Here 1 3

n 2, p ,q 1 p4 4

. [1/2]

X 0 1 2 Total P(X) 0 2

20

1 3 9C

4 4 16

1 2 1

21

1 3 6C

4 4 16

2 2 2

22

1 3 1C

4 4 16

X P(X)

0 6

16

2

16

8

16

2X P(X)

0 6

16

4

16

10

16

Mean, 8 1

E(x) X P(X) or16 2

, [1/2]

Var (X) 2 2 10 1 6 3X P(X) (Mean) or

16 4 16 8 . [1]

SECTION D Q24. Let y f (x) x [0, )

2 2y 9x 6x 5 (3x 1) 6

Clearly when x [0, ) then y 5 Range of f [ 5, ) Codomain of f

Hence f is not onto and, therefore f (x) is not invertible. [2] Let us take the modified codomain of f as [ 5, ) . [1/2]

Let’s now check whether f is one-one : Let 1 2x , x [0, ) such that 1 2f (x ) f (x )

2 21 2(3x 1) 6 (3x 1) 6 1 2 1 23x 1 3x 1 x x . So, f is one-one.

Since, with the modified codomain, we have range of f = new codomain of f. So, f is onto. Hence f is bijective function and, therefore f is invertible. [1+1/2]

Now 2for any y [ 5, ), y (3x 1) 6

2y 6 (3x 1) y 6 1

x3

[1]

1 1 y 6 1f :[ 5, ) [0, ), f (y)

3

. [1]

OR Let A Q Q , where Q is the set of all rational numbers, and * be a binary operation on

A defined by (a, b)*(c,d) (ac, b ad) for (a, b), (c,d) A .

Note that (1,2)*(2,3) (2,5) and (2,3)*(1, 2) (2,7) which implies (2,5) (2,7)

That is, (a, b)*(c,d) (c,d)*(a, b) (a,b),(c,d) Q Q . So, * isn’t commutative. [1]

Let (a, b), (c,d), (e, f ) Q Q .

Now [(a, b)*(c,d)]*(e, f ) (ac,b ad)*(e, f ) (ace,b ad acf )...(i)

Also, (a, b)*[(c,d)*(e, f )] (a,b)*(ce,d cf ) (ace, b ad acf )...(ii)

By (i) and (ii), [(a,b)*(c,d)]*(e, f ) (a, b)*[(c,d)*(e, f )] .

Hence * is associative. [1]

[1+1/2] [1/2]



Let (e,e ) be the identity element of * in A. Then (a, b)*(e,e ) (a,b) (e,e )*(a, b)

(ae,b ae ) (a,b) ae a e 1

Identity element (1,0)b ae b e 0

. [2]

Let (x, y) be the inverse of an element (a, b) of A. Then (a, b)*(x, y) (1,0) (x, y)*(a,b)

(ax,b ay) (1,0)

1ax 1 x

a

bb ay 0 y

a

1 b

Inverse of (a, b) ,a a

. [2]

Q25. LHS : Let

2

2

2

(a b)c c

c

(b c)a a

a

(c a)b b

b

By 1 1 2 2 3 3R cR , R aR , R bR ,

2 2 2

2 2 2

2 2 2

(a b) c c1

a (b c) aabc

b b (c a)

[1/2]

By 1 1 3 2 2 3C C C and C C C ,

2

2

2

(a b c)(a b c) 0 c1

0 (b c a)(b c a) aabc

(b c a)(b c a) (b c a)(b c a) (c a)

[1]

Taking (a + b + c) common from C1 and C2 both

2

22

2

(a b c) 0 c(a b c)

0 (b c a) aabc

(b c a) (b c a) (c a)

[1/2]

By 3 3 1 2R R (R R ) ,

2

22

(a b c) 0 c(a b c)

0 (b c a) aabc

2a 2c 2ca

[1]

By 1 1 2 2C cC , C aC

2 2

22 2

(ac bc c ) 0 c(a b c)

0 (ba ca a ) aabcac

2ca 2ca 2ca

[1/2]

By 1 1 3 2 2 3C C C , C C C

2 2

22 2

ac bc c c(a b c)

a ba ca aabcac

0 0 2ca

[1]

Take c, a, 2ca common from 1 2 3R , R and R ,



2 2 2

a b c c2c a (a b c)

a b c aabcac

0 0 1

[1/2]

Expanding along R3,

2

22(a b c)0 0 1 ab ac b bc ac

b

2

2 22(a b c)ab b bc 2(a b c) RHS

b

. [1]

OR We have

p q p q

q r q r 0

p q q r 0

By 1 1R R ,

2p q p q1

q r q r 0

p q q r 0

[1]

By 3 3 1 2R R (R R ) ,

2

2

p q p q1

q r q r 0

0 0 p 2q r

[2]

Expanding along R1, 2 210 0 p 2q r p r q 0

[1]

2 21(q pr) 2q r p 0

[1]

Either 2 2q pr 0 or 2q r p 0 . That implies, 2 2q pr or p( ) 2q( ) r 0

Therefore either p, q, r are in G.P. or, satisfies the equation 2px 2qx r 0 .

Hence either p, q, r are in G.P. or, is a root of the equation 2px 2qx r 0 . [1]

Q26. We have 2y 5 x … (i) and x 1, if x 1

y x 11 x, if x 1

...(ii)

Solving (i) and (ii), 2 2 2| x 1| [ 5 x ]

2 2x 2x 1 5 x 2x x 2 0

(x 1)(x 2) 0 x 1, 2 [1]

2 2

i ii

1 1

Required area y dx y dx

2 2

2

1 1

5 x dx | x 1| dx

[2]

2 1 2

2

1 1 1

5 x dx | x 1| dx | x 1| dx

2 1 2

2

1 1 1

5 x dx (x 1)dx (x 1)dx

1 22 2 22 1

1 1 1

x 5 x (x 1) (x 1)5 x sin

2 2 2 25

y = |x – 1|

2y 5 x

5, 0

(1, 0) (–1, 0) [1 Mark for correct Figure] (2, 0)



1 22 2 22 1

1 1 1

x 5 x (x 1) (x 1)5 x sin

2 2 2 25

[1+1/2]

1 15 2 5 1 11 sin 1 sin 0 2 0

2 2 25 5

1 15 2 1 1sin sin sq.units

2 25 5

[1/2]

15 2 1 1 4 1sin 1 1

2 5 5 25 5

15 4 1 1 5 1sin

2 5 5 2 2 2 2

5 1sq.units

4 2

.

Q27. Let 2

4 4

0

x sin x cos xI dx

sin x cos x

…(i)

2

4 40

x sin x cos x2 2 2

I dx

sin x cos x2 2

2

4 4

0

x cos x sin x2

I dxcos x sin x

…(ii) [1]

Adding (i) and (ii), 2

4 4

0

cos x sin x2I dx

2 cos x sin x

[1/2]

4

4 4

0

cos x sin x2I 2 dx

2 cos x sin x

4 2

4 4 4 4

0

4

cos x sin x cos x sin x2I dx dx

2 cos x sin x cos x sin x

2 24 2

4 4

0

4

tan x sec x cot x cosec x2I dx dx

2 1 tan x cot x 1

[2]

Substituting 2 2tan x t and cot x u in both integrals respectively

2 22 tan x sec xdx dt and 2cosec x cot xdx du

Also when x 0 t 0, when x t 14

And, when x u 1, when x u 04 2

[1]

1 0

2 2

0 1

1 dt 1 du2I

2 2 1 t 2 1 u

1 1

2 2

0 0

1 dt 1 du2I

2 2 1 t 2 1 u

Give full marks whether this is shown or not.



1 1 1 10 0I [tan t] [tan u]

8

[1]

2

I8 4 4 16

or, 2

I4

. [1/2]

OR 0

42xLet I (x e ) dx

b

h 0a

f (x)dx lim h f (a) f (a h) f (a 2h) ... f (a (n 1)h)

b n 1

nr 0a

f x dx lim h f (a rh)

n , h 0 nh b a

Here 2xf (x) x e and a 0, b 4 so, nh 4 0 4 . [1]

2rh2 0 2rhf (0 rh) 0 rh e rh e

n 1

2rh

nr 00

42xI (x e ) dx lim h rh e

[1]

n 1 n 1

2rh

nr 0 r 0

I lim h h r e

2h 4h 2(n 1)h

n

n(n 1)I lim h h 1 e e ... e

2

2h n

2hn

nh(nh h) 1 [(e ) 1]I lim h

2 e 1

[2]

2nh

2hn

nh(nh h) 2h e 1I lim

2 e 1 2

[1]

2 44(4 0) e 1

I 12 2

815 e

I2

. [1]

Q28. The normal vectors to the planes x y 2z 3 and 2x y 3z 0 are respectively ˆ ˆ î j 2k

and ˆ ˆ ˆ2i j 3k .

Now 1 2

ˆ ˆ î j kˆ ˆ ˆm m m 1 1 2 5i 7 j k

2 1 3

. [2]

The equation of required plane through (4, –3, 2) is, ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆr.(5i 7 j k) (4i 3j 2k).(5i 7 j k)

That is, ˆ ˆ ˆr.(5i 7 j k) 1

…(i) is the equation of required plane. [1]

Now for the point of intersection of plane (i) and line ˆ ˆ ˆ ˆ ˆ ˆr i 2 j k (i 3j 9k)

, we have

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ î 2j k (i 3j 9k) .(5i 7 j k) 1 5 14 1 (5 21 9) 1 1 [1+1+1/2]

So, position vector of the required point is ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆr i 2j k ( 1)(i 3j 9k) 8k j

. [1/2]

So the point of intersection si (0, –1, 8). Q29. Let x kg of Food 1 be mixed with y kg of Food 2. To minimize Z = ` (50x + 70y) [1]



Subject to the constraints : 2x y 8, x 2y 10,x 0, y 0 [1]

Corner Points Value of Z (in `)

A(0, 8) 560 B(2, 4) 380 Min. value

C(10, 0) 500 Since feasible region is unbounded so, 380 may or may not be minimum value of z. To check, draw 50x 70y 380 i.e., 5x 7y 38 .

As in the half plane 5x 7y 38 , there is no point

common with the feasible region. Hence minimum value of Z is ` 380. [1]

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# This paper has been issued by CBSE, New Delhi for 2017 Board Exams of XII. Note : We’ve re-typed the same and done the necessary corrections (as the original paper had some errors). On other hand, if you still find any error which could have gone un-noticed, please do inform us via WhatsApp @ +919650350480 or Email us : [email protected]

B

5x + 7y = 38 [2 Marks for Graph]

[1]

mathematics - theopgupta.com · ... b) * (c, d) = (ac, b + ad), where q is ... find the identity...

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